C. The force of air resistance acts in the same direction as the force of gravity
Answer:
See attached picture!
Explanation:
There are two types of radioactive decay: alpha and beta.
Alpha decay removes a helium atom from the original element, causing the original mass number to decrease by 4 and the original atomic number to decrease by 2.
Beta decay removes an electron from the original element, causing the original mass number to remain the same and the original atomic number to increase by 1.
To solve your problems, you need to identify which decay is most likely to happen. You could do this by checking the possible answer choices. Usually, professors tell you which decay you are expected to write an equation of, but in this case, it's a matter of trial and error in order to get the answer.
Answer : The current passing between the electrodes is, 
Explanation :
First we have to calculate the charge of sodium ion.

where,
q = charge of sodium ion
n = number of sodium ion = 
e = charge on electron = 
Now put all the given values in the above formula, we get:

Now we have to calculate the charge of chlorine ion.

where,
q' = charge of chlorine ion
n = number of chlorine ion = 
e = charge on electron = 
Now put all the given values in the above formula, we get:

Now we have to calculate the current passing between the electrodes.



Thus, the current passing between the electrodes is, 
Answer:
yes
Explanation:
Usually, it would not affect the crucible, but depending on the temperature of the flame the enamel of the crucible may begin to melt and stick to the metal object being used to handle the crucible. This tiny amount that is melted off can cause very small changes in the original mass of the crucible, which although it is almost unnoticeable it is still there. Therefore, the answer to this question would be yes.
Anions are solvated in protic hydrogen-bonding solvents (such as ethanol). Consequently, nucleophiles are less reactive. Since soft nucleophiles are less strongly solvated than hard nucleophiles, these solvents boost the relative reactivity of soft anions.
<h3>
Ethanol is either a nucleophile or a base.</h3>
The ethanol is a base Because carbocation is an extremely reactive species, a base or nucleophile as weak as ethanol can replace or remove it. SN1 and E1 would not be conceivable without the carbocation or a strong departing group.
<h3>How do solvents impact anionic nucleophile's reactivity?</h3>
In polar aprotic solvents, nucleophilic substitution reactions of anionic nucleophiles often proceed more quickly. The normal relative reactivity order in such solvents (like DMSO)is Anions are solvated in protic hydrogen-bonding solvents (such as ethanol). Consequently, nucleophiles are less reactive.
Learn more about nucleophiles here:-
brainly.com/question/27127109
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